Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 2x}{x - 6} = \dfrac{x + 18}{x - 6}$
Multiply both sides by $x - 6$ $ \dfrac{x^2 - 2x}{x - 6} (x - 6) = \dfrac{x + 18}{x - 6} (x - 6)$ $ x^2 - 2x = x + 18$ Subtract $x + 18$ from both sides: $ x^2 - 2x - (x + 18) = x + 18 - (x + 18)$ $ x^2 - 2x - x - 18 = 0$ $ x^2 - 3x - 18 = 0$ Factor the expression: $ (x - 6)(x + 3) = 0$ Therefore $x = 6$ or $x = -3$ At $x = 6$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 6$, it is an extraneous solution.